Graphs: Sequences, Functions and Graphs

Q: How do I calculate the gradient of a straight line from two points?

Use the formula where and are the coordinates of the two points. A line going up left to right has positive gradient, going down has negative gradient.

Q: How do I solve equations using graphs?

To solve , plot and . The x-values of their intersection points are the solutions, which correspond to roots of .

Edexcel International GCSE Mathematics A· 3.3· 25 min read

1. Cartesian Coordinates and Midpoints★☆☆☆☆⏱ 5 min

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📘 Definition

Cartesian Coordinates

A system for locating points on a plane using two perpendicular axes: the horizontal x-axis and vertical y-axis. Points are written as $(x, y)$ where x is the horizontal position and y is the vertical position.

📐 Worked Example

Find the midpoint of the line segment connecting points $A (2, 5)$ and $B (- 4, 1)$ .

1
Recall the midpoint formula:
$\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$
2
Substitute x values to find the x-coordinate of the midpoint:
$\frac{2 + (-4)}{2} = -1$
3
Substitute y values to find the y-coordinate of the midpoint:
$\frac{5 + 1}{2} = 3$
4
The midpoint of segment AB is $(- 1, 3)$

✓ Quick check

What are the coordinates of the point 3 units left and 2 units up from $(1, - 2)$ ?
- A) $(- 2, 0)$
- B) $(4, - 4)$
- C) $(- 2, - 4)$
- D) $(4, 0)$
Reveal answer
A) $(-2, 0)$ —
Moving left reduces the x value by 3 ( $1 - 3 = - 2$ ), moving up increases the y value by 2 ( $- 2 + 2 = 0$ ).

Exam tip:

Always plot the x-coordinate first, then the y-coordinate: remember x comes before y in the alphabet to avoid mixing up values.

2. Straight Line Graphs and Gradient★★☆☆☆⏱ 7 min

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📘 Definition

Gradient

The steepness of a straight line, calculated as the change in y (vertical rise) divided by the change in x (horizontal run) between two points on the line, denoted $m$ .

📐 Worked Example

Find the equation of the straight line passing through $(0, 3)$ with gradient 2, then rewrite it in the form $a x + b y = c$ .

1
Recall the straight line equation $y = m x + c$ , where $c$ is the y-intercept (when $x = 0$ , $y = 3$ so $c = 3$ , and $m = 2$ )
2
Write the equation in slope-intercept form:
$y = 2x + 3$
3
Rearrange to standard form by moving all terms to one side:
$2x - y = -3$

📐 Worked Example

Find the equation of the line perpendicular to $y = 2 x + 3$ that passes through $(4, 1)$ .

1
The gradient of the given line is 2, so the perpendicular gradient is the negative reciprocal: $m = - \frac{1}{2}$
2
Substitute $m = - 0.5$ , $x = 4$ and $y = 1$ into $y = m x + c$ to solve for $c$ :
$1 = (-0.5 \times 4) + c \implies 1 = -2 + c \implies c = 3$
3
The equation of the perpendicular line is $y = - 0.5 x + 3$

Exam tip:

For lines given in $a x + b y = c$ form, rearrange to $y = m x + c$ to quickly find the gradient and y-intercept instead of calculating from two points.

3. Interpreting Standard Real-World Graphs★★☆☆☆⏱ 5 min

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Distance-time graphs: Gradient = speed, flat sections = stationary
Speed-time graphs: Gradient = acceleration, area under line = total distance travelled
Conversion graphs: Used to convert between two units (e.g. currency, km to miles) by reading values off the straight line

📐 Worked Example

A distance-time graph shows a car travelling 80km in 2 hours, stopping for 1 hour, then returning 80km in 2.5 hours. Calculate the average speed for the entire journey, excluding the stop.

1
Calculate total distance travelled: $80 + 80 = 160$ km
2
Calculate total moving time: $2 + 2.5 = 4.5$ hours
3
Average speed = total distance / total moving time:
$\frac{160}{4.5} \approx 35.6 \text{ km/h}$

Exam tip:

Always check the units on the axes of real-world graphs to avoid unit conversion errors in calculations.

4. Non-Linear Graphs★★★☆☆⏱ 6 min

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📐 Worked Example

Complete the table for $y = x^{2} - 2 x - 3$ for x from -1 to 4, then describe the graph shape.

1
Substitute each x value into the equation to find y: $x = - 1 ⟹ y = 0$ , $x = 0 ⟹ y = - 3$ , $x = 1 ⟹ y = - 4$ , $x = 2 ⟹ y = - 3$ , $x = 3 ⟹ y = 0$ , $x = 4 ⟹ y = 5$
2
The graph is a symmetric U-shaped curve called a parabola, with a minimum turning point at $(1, - 4)$

📐 Worked Example

Find the gradient of the curve $y = x^{2}$ at $x = 2$ using a tangent.

1
Draw a straight tangent line that touches the curve only at the point $(2, 4)$ , balanced evenly on either side of the point
2
Pick two easy-to-read points on the tangent, e.g. $(1, 1)$ and $(3, 7)$
3
Calculate the gradient of the tangent:
$\frac{7 - 1}{3 - 1} = \frac{6}{2} = 3$
4
The gradient of the curve at $x = 2$ is 3

Exam tip:

When plotting non-linear graphs, always draw a smooth continuous curve through points: never draw straight line segments between points for quadratic, cubic or reciprocal graphs.

5. Function Transformations and Graphical Solutions★★★★☆HL only⏱ 7 min

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📘 Definition

Function Transformation

A change to the graph of a function $f (x)$ that shifts, stretches or reflects the graph without changing its core shape. Only four transformations are required for this exam.

📐 Worked Example

The graph of $f (x) = x^{2}$ is transformed to give $g (x) = (x - 3)^{2} + 2$ . Describe the full transformation.

1
$f (x + a)$ shifts the graph left by $a$ units, so $f (x - 3)$ is a shift right by 3 units
2
$f (x) + a$ shifts the graph up by $a$ units, so adding 2 is a shift up by 2 units
3
The full transformation is a translation by the vector $(32)$

📐 Worked Example

Use the graphs of $y = x^{2}$ and $y = x + 2$ to solve the equation $x^{2} - x - 2 = 0$ .

1
Rearrange the target equation to match the graph forms: $x^{2} - x - 2 = 0 ⟹ x^{2} = x + 2$
2
The solutions are the x-values of the intersection points of the two graphs, which are $x = - 1$ and $x = 2$

Exam tip:

When describing transformations, always test key points (e.g. turning points, intercepts) to confirm you have applied the shift direction correctly: $f (x + a)$ shifts left, not right.

6. Common Pitfalls

Wrong move:

Plotting $(y, x)$ instead of $(x, y)$ for coordinates

Why:

Confusing the order of x and y values

Correct move:

Remember x comes before y in the alphabet, so always plot the horizontal position first, then the vertical position.

Wrong move:

Calculating gradient as $\frac{Δ x}{Δ y}$ instead of $\frac{Δ y}{Δ x}$

Why:

Mixing up numerator and denominator in the gradient formula

Correct move:

Use the mnemonic 'rise over run': rise is vertical (y) change, run is horizontal (x) change, so gradient = rise / run.

Wrong move:

Assuming $f (x + a)$ shifts the graph to the right

Why:

Associating positive a values with rightward movement

Correct move:

$f (x + a)$ shifts the graph left by a units, $f (x - a)$ shifts right by a units: test with the x=0 position of key points to confirm.

Wrong move:

Drawing straight line segments between points on non-linear graphs

Why:

Rushing plotting without checking the function shape

Correct move:

For quadratic, cubic or reciprocal graphs, draw a smooth, continuous curve through all plotted points with no sharp corners.

Wrong move:

Using only the negative of a gradient for perpendicular lines instead of the negative reciprocal

Why:

Simplifying the perpendicular gradient rule incorrectly

Correct move:

If the original gradient is m, the perpendicular gradient is $- \frac{1}{m}$ : multiply the two gradients to confirm they equal -1 before proceeding.

7. Quick Reference Cheatsheet

Concept	Foundation Tier	Higher Tier (Addition)
Coordinates	Plot $(x, y)$ in 4 quadrants, midpoint formula $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$	Same as Foundation
Straight Lines	Gradient = $\frac{Δ y}{Δ x}$ , equation $y = m x + c$ , rearrange to $a x + b y = c$	Parallel lines: same $m$ , perpendicular lines: $m_{1} \times m_{2} = - 1$
Graph Types	Linear, quadratic, distance/time, speed/time, conversion graphs	Cubic, reciprocal, $y = sin x$ , $y = cos x$ , $y = tan x$ (degrees only)
Gradients	Calculate straight line gradients only	Find non-linear gradients by drawing a tangent to the curve
Transformations	Not required	Four transformations: $f (x) + a$ (shift up/down), $f (x + a)$ (shift left/right), $a f (x)$ (vertical stretch), $f (a x)$ (horizontal stretch)
Graphical Solutions	Not required	Intersection of $y_{1}$ and $y_{2}$ gives solutions to $y_{2} - y_{1} = 0$

8. Frequently Asked

How do I calculate the gradient of a straight line from two points?

Use the formula $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ where $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are the coordinates of the two points. A line going up left to right has positive gradient, going down has negative gradient.

What is the rule for perpendicular line gradients?

If two lines are perpendicular, the product of their gradients equals -1. If a line has gradient $m$ , its perpendicular has gradient $- \frac{1}{m}$ . Horizontal and vertical lines are always perpendicular as an exception.

How do I solve equations using graphs?

To solve $f (x) = m x + c$ , plot $y = f (x)$ and $y = m x + c$ . The x-values of their intersection points are the solutions, which correspond to roots of $f (x) - m x - c = 0$ .

Going deeper

What's Next

Now that you have mastered graph content for Edexcel IGCSE Math A, you can move on to Higher tier calculus, where you will learn to calculate non-linear gradients algebraically using differentiation. You should also practice applying graph skills to real-world problem solving questions, which frequently combine graph interpretation with algebra and arithmetic in both Foundation and Higher papers. Make sure to practice past paper graph questions to familiarise yourself with marking criteria, especially for plotting questions where correct axis labels and smooth curves are required for full marks.